#include <iostream>
#include <vector>
#include <algorithm>
#include <climits>

using namespace std;

int minNap(int n, int p, int db, int ossz, const vector<int>& v) {
    // Base case: If the target is reached, return the current day count
    if (ossz >= p) {
        return db;
    }
    // Base case: If no more elements are left, return an impossible value
    if (n < 0) {
        return INT_MAX;
    }
    // Recursive step: Explore both including and excluding the current day
    int include = minNap(n - 1, p, db + 1, ossz + v[n], v); // Include the current day's work
    int exclude = minNap(n - 1, p, db, ossz, v);           // Skip the current day

    // Return the minimum of the two results
    return min(include, exclude);
}

int main() {
    int n, p;
    cin >> n >> p;
    vector<int> v(n);
    for (auto& i : v)
        cin >> i;

    int result = minNap(n - 1, p, 0, 0, v); // Start from the last day
    if (result == INT_MAX) {
        cout << -1 << endl; // Output -1 if no valid solution exists
    } else {
        cout << result << endl; // Output the minimal number of days
    }
    return 0;
}